本次笔记均为OCR结果，仅供参考

均匀平面波在介质面的反射和折射

(入射波的波矢 ki⃗\vec{k_i} 位于 yOz 平面内)

入射波: Ei⃗=E0i⃗eiki⃗⋅r⃗=E0i⃗ei(kiyy+kizz)\vec{E_i} = \vec{E_{0i}} e^{i \vec{k_i} \cdot \vec{r}} = \vec{E_{0i}} e^{i(k_{iy} y + k_{iz} z)}

反射波: Er⃗=E0r⃗eikr⃗⋅r⃗=E0r⃗ei(krxx+kryy+krzz)\vec{E_r} = \vec{E_{0r}} e^{i \vec{k_r} \cdot \vec{r}} = \vec{E_{0r}} e^{i(k_{rx} x + k_{ry} y + k_{rz} z)}

折射波: Et⃗=E0t⃗eikt⃗⋅r⃗=E0t⃗ei(ktxx+ktyy+ktzz)\vec{E_t} = \vec{E_{0t}} e^{i \vec{k_t} \cdot \vec{r}} = \vec{E_{0t}} e^{i(k_{tx} x + k_{ty} y + k_{tz} z)}

介质 1 侧的总电磁场: E1⃗=Ei⃗+Er⃗\vec{E_1} = \vec{E_i} + \vec{E_r}

介质 2 侧的总电磁场: E2⃗=Et⃗\vec{E_2} = \vec{E_t}

1. 方向关系: 由边值关系 n^×(E2⃗−E1⃗)=0\hat{n} \times (\vec{E_2} - \vec{E_1}) = 0 可知 (电场切向相等)

n^×(Ei⃗+Er⃗)∣z=0=n^×Et⃗∣z=0\hat{n} \times (\vec{E_i} + \vec{E_r})|_{z=0} = \hat{n} \times \vec{E_t}|_{z=0}

n^×(E0i⃗ei(kiyy)+E0r⃗ei(krxx+kryy))=n^×E0t⃗ei(ktxx+ktyy)\hat{n} \times (\vec{E_{0i}} e^{i(k_{iy} y)} + \vec{E_{0r}} e^{i(k_{rx} x + k_{ry} y)}) = \hat{n} \times \vec{E_{0t}} e^{i(k_{tx} x + k_{ty} y)}（在界面上任意位置成立）

则: krx=ktx=0k_{rx} = k_{tx} = 0

Acos⁡kiyy+Bcos⁡(krxx+kryy)=Ccos⁡(ktxx+ktyy)A \cos k_{iy} y + B \cos (k_{rx} x + k_{ry} y) = C \cos (k_{tx} x + k_{ty} y)

那么, 入射波、反射波、折射波共面, kiy=kry=ktyk_{iy} = k_{ry} = k_{ty}

kisin⁡θi=krsin⁡θr=ktsin⁡θtk_i \sin \theta_i = k_r \sin \theta_r = k_t \sin \theta_t

即 ki=krk_i = k_r, θi=θr\theta_i = \theta_r, sin⁡θisin⁡θt=ktki=ωμ2ϵ2ωμ1ϵ1=μ2ϵ2μ1ϵ1=n2n1\frac{\sin \theta_i}{\sin \theta_t} = \frac{k_t}{k_i} = \frac{\omega \sqrt{\mu_2 \epsilon_2}}{\omega \sqrt{\mu_1 \epsilon_1}} = \sqrt{\frac{\mu_2 \epsilon_2}{\mu_1 \epsilon_1}} = \frac{n_2}{n_1} (折射定律)

2. 幅度关系:

分类讨论: (1) N 波: E⃗\vec{E} 垂直于入射面

{n^×(E2⃗−E1⃗)=0n^×(H2⃗−H1⃗)=0\begin{cases} \hat{n} \times (\vec{E_2} - \vec{E_1}) = 0 \\ \hat{n} \times (\vec{H_2} - \vec{H_1}) = 0 \end{cases}

{E0i⃗+E0r⃗=E0t⃗E0iη1cos⁡θi−E0rη1cos⁡θi=E0tη2cos⁡θt\begin{cases} \vec{E_{0i}} + \vec{E_{0r}} = \vec{E_{0t}} \\ \frac{E_{0i}}{\eta_1} \cos \theta_i - \frac{E_{0r}}{\eta_1} \cos \theta_i = \frac{E_{0t}}{\eta_2} \cos \theta_t \end{cases}

E0rE0i=η2cos⁡θi−η1cos⁡θtη2cos⁡θi+η1cos⁡θt≈μr∼1n1cos⁡θi−n2cos⁡θtn1cos⁡θi+n2cos⁡θt=−sin⁡(θi−θt)sin⁡(θi+θt)\frac{E_{0r}}{E_{0i}} = \frac{\eta_2 \cos \theta_i - \eta_1 \cos \theta_t}{\eta_2 \cos \theta_i + \eta_1 \cos \theta_t} \stackrel{\mu_r \sim 1}{\approx} \frac{n_1 \cos \theta_i - n_2 \cos \theta_t}{n_1 \cos \theta_i + n_2 \cos \theta_t} = - \frac{\sin (\theta_i - \theta_t)}{\sin (\theta_i + \theta_t)}

\frac{E_{0t}}{E_{0i}} = \frac{2 \eta_2 \cos \theta_i}{\eta_2 \cos \theta_i + \eta_1 \cos \theta_t} \stackrel{\mu_t \sim 1}{\approx} \frac{2 n_1 \cos \theta_i}{n_1 \cos \theta_i + n_2 \cos \theta_t} = \frac{2 \cos \theta_i \sin \theta_t}{\sin (\theta_i + \theta_t)} $$ (菲涅尔公式) (2) P 波: $\vec{E}$ 平行于入射面:

\frac{E_{0r}}{E_{0i}} = \frac{\tan (\theta_i - \theta_t)}{\tan (\theta_i + \theta_t)}

\frac{E_{0t}}{E_{0i}} = \frac{2 \cos \theta_i \sin \theta_t}{\sin (\theta_i + \theta_t) \cos (\theta_i - \theta_t)}

### 关于反射波的结论: (1) $n_1 < n_2$ 时, 光疏 → 光密 N 波 $sgn \{ \frac{E_{0r}}{E_{0i}} \} = -1$ 存在半波损失 (相位变化 π) (2) $n_1 > n_2$ 时, 光密 → 光疏. $\sin \theta_i / \sin \theta_t = n_2 / n_1 = n_{21} < 1$ 则 $\sin \theta_t = \sin \theta_i / n_{21}$ $\theta_i \in [0, \frac{\pi}{2}]$ 时 当 $\theta_i > \theta_c = \arcsin (n_{21})$ 时, $\sin \theta_t > 1$ 发生全反射 (3) 当 $\theta_i + \theta_t = \pi / 2$ 时, $\theta_i = \theta_B = \arctan (\frac{n_2}{n_1})$ 布儒斯特角 ### 全反射现象 $\sin \theta_t = \sin \theta_i / n_{21} > 1$ 且 $\theta_t$ 是复数. $\cos \theta_t = \sqrt{1 - \sin^2 \theta_t} = i \sqrt{\sin^2 \theta_i / n_{21}^2 - 1}$ 反射波幅度改变: N 波

\frac{E_{0r}}{E_{0i}} = \frac{\cos \theta_i - i \sqrt{\frac{\sin^2 \theta_i}{n_{21}^2} - 1}}{\cos \theta_i + i \sqrt{\frac{\sin^2 \theta_i}{n_{21}^2} - 1}} = e^{-i 2 \phi_\pi}

P波 P 波

\frac{E_{0r}}{E_{0i}} = \frac{n_{21}^2 \cos \theta_i - i \sqrt{\sin^2 \theta_i - n_{21}^2}}{n_{21}^2 \cos \theta_i + i \sqrt{\sin^2 \theta_i - n_{21}^2}} = e^{-i 2 \phi_P}

一般 $\phi_\pi \ne \phi_P$ ### 介质1侧总电磁场 以N波为例：

\vec{E}1 = \vec{E}i + \vec{E}r = \vec{E}{0i} e^{i(k{iy} y + k{iz} z)} + \vec{E}{0r} e^{i(k{ry} y + k_{rz} z)}

$k_{iy} = k_{ry} = k_y,k_{iz} = -k_{rz},\vec{E}_{0i} = \vec{E}_{0}$ $\vec{E}_{0r} = \vec{E}_{0} e^{-i 2 \phi_{N}}$

\vec{E}i = \vec{E}{i0} e^{i(k_{ix}x + k_{iy}y + k_{iz}z)}

\vec{E}r = \vec{E}{r0} e^{i(k_{rx}x + k_{ry}y + k_{rz}z)}

\vec{E} = \vec{E}i + \vec{E}r = \vec{E}{i0} e^{i(k{ix}x + k_{iy}y + k_{iz}z)} + \vec{E}{r0} e^{i(k{rx}x + k_{ry}y + k_{rz}z)}

\begin{aligned}
\vec{E}1 &= \vec{E}0 e^{i(k{y}y + k{iz}z)} + \vec{E}{0} e^{-i 2 \phi{N}} e^{i(k_{y}y - k_{z}z)} \
&= 2 \cos (k_z z + \phi_{N}) \vec{E}{0} e^{-i\phi_N} e^{i k{y} y} \
\end{aligned}

结果的前半部分沿z轴周期性变化，为复振幅。后半部分说明电磁波相位传播沿y方向相速度：结果的前半部分沿z轴周期性变化，为复振幅。后半部分说明电磁波相位传播沿y方向 相速度：

V_P = \frac{\omega}{k_y} =\frac{\omega}{k_i \sin \theta_i} > \frac{\omega}{k_i} = \frac{c}{n}

称为快波 介质2侧的电磁场（要满足边界条件，必然不是全都没有） $$ \vec{E_t} = \vec{E_{ot}} e^{i(k_{ty} y + k_{tz} z)}

$ k_{ty} = k_t \sin \theta_t = k_i \sin \theta_i k_{tz} = k_t \cos \theta_t = ik_i \sqrt{\sin^2 \theta_i - n_{21}^2} = i K_{tz} $

所以 Et⃗=Eot⃗e−Ktzzeiktyy\vec{E_t} = \vec{E_{ot}} e^{-K_{tz} z} e^{i k_{ty} y}
依然能够分为沿z轴的复振幅和沿y轴的相位传播

\quad 倏逝波/消逝波

相速度 vp=ωkty=ωktsin⁡θt<ωkt=cn2v_p = \frac{\omega}{k_{ty}} = \frac{\omega}{k_t \sin \theta_t} < \frac{\omega}{k_t} = \frac{c}{n_2} 称作慢波

穿透深度： δ=1Ktz=λ12πsin⁡2θi−n212\delta = \frac{1}{K_{tz}} = \frac{\lambda_1}{2\pi \sqrt{\sin^2 \theta_i - n_{21}^2}} \quad ←\leftarrow 介质1波长

幅度：
\quad S波： EotEoi=2cos⁡θicos⁡θi+isin⁡2θi−n212\frac{E_{ot}}{E_{oi}} = \frac{2 \cos \theta_i}{\cos \theta_i + i \sqrt{\sin^2 \theta_i - n_{21}^2}}

\quad P波： EotEoi=2n21cos⁡θin212cos⁡θi+isin⁡2θi−n212\frac{E_{ot}}{E_{oi}} = \frac{2 n_{21} \cos \theta_i}{n_{21}^2 \cos \theta_i + i \sqrt{\sin^2 \theta_i - n_{21}^2}}

能流： Si⃗\vec{S_i} \quad N波为例

从波印廷矢量：
\quad 入射波复数波印廷矢量： Si~=12Ei×Hi∗=∣Eoi∣2ki^2η1\tilde{S_i} = \frac{1}{2} E_i \times H_i^* = |E_{oi}|^2 \frac{\hat{k_i}}{2\eta_1}
\quad 反射波复数波印廷矢量： Sr~=12η1∣Eor∣2kr^=sin⁡2(θi−θt)sin⁡2(θi+θt)∣Eoi∣22ηikt^\tilde{S_r} = \frac{1}{2\eta_1} |E_{or}|^2 \hat{k_r} = \frac{\sin^2(\theta_i - \theta_t)}{\sin^2(\theta_i + \theta_t)} \frac{|E_{oi}|^2}{2\eta_i} \hat{k_t}
\quad 折射波 St~=12η2∣Eot∣2kt^=4cos⁡2θisin⁡2θtsin⁡2(θi+θt)12η2∣Eoi∣2kt^\tilde{S_t} = \frac{1}{2\eta_2} |E_{ot}|^2 \hat{k_t} = \frac{4\cos^2\theta_i \sin^2\theta_t}{\sin^2(\theta_i + \theta_t)} \frac{1}{2\eta_2} |E_{oi}|^2 \hat{k_t}

在垂直界面方向(法向) n^⋅⟨Si⃗⟩+n^⋅⟨Sr⃗⟩=n^⋅⟨St⃗⟩\hat{n} \cdot \langle \vec{S_i} \rangle + \hat{n} \cdot \langle \vec{S_r} \rangle = \hat{n} \cdot \langle \vec{S_t} \rangle

功率反射率： R=n^⋅⟨Sr⃗⟩n^⋅⟨Si⃗⟩=∣EorEoi∣2R = \frac{\hat{n} \cdot \langle \vec{S_r} \rangle}{\hat{n} \cdot \langle \vec{S_i} \rangle} = \left| \frac{E_{or}}{E_{oi}} \right|^2

功率透射率： T=n^⋅⟨St⃗⟩n^⋅⟨Si⃗⟩=∣EotEoi∣22η1cos⁡θtη2cos⁡θiT = \frac{\hat{n} \cdot \langle \vec{S_t} \rangle}{\hat{n} \cdot \langle \vec{S_i} \rangle} = \left| \frac{E_{ot}}{E_{oi}} \right|^2 \frac{2\eta_1 \cos \theta_t}{\eta_2 \cos \theta_i}

R+T=1R+T=1 \quad 全反射时 n^⋅⟨Si⃗⟩=−n^⋅⟨Sr⃗⟩\hat{n} \cdot \langle \vec{S_i} \rangle = - \hat{n} \cdot \langle \vec{S_r} \rangle
n^⋅⟨St⃗⟩=0\qquad \qquad \hat{n} \cdot \langle \vec{S_t} \rangle = 0
n^⋅St~=∣Eot∣22η2cos⁡θt\qquad \qquad \hat{n} \cdot \tilde{S_t} = \frac{|E_{ot}|^2}{2\eta_2} \cos \theta_t 纯虚数

介质2侧沿垂直于界面方向没有平均的能流，只有瞬时能流。